Title: Proof Methods: Part 2
1Proof Methods Part 2
2More Number Theory Definitions
- Divisibility
- n is divisible by d iff
- ?k?Z ndk
- dn is read d divides n where n and d are
integers and d ? 0 (note dn ? d/n) - Other ways to say it
- n is a multiple of d
- d is a factor of n
- d is a divisor of n
- d divides n
- Properties of Divisibility
- Thm 3.3.1 Transitivity
- ?a,b,c ?Z, If a divides b and b divides c, then a
divides c - Thm 3.3.2 Divisibility by a Prime
- Any integer ngt1 is divisible by a prime number
3Transitivity Proof
- Prove that for all integers a, b, and c, If a
divides b and b divides c, then a divides c - ?a,b,c ?Z, If ab and bc then ac
- Suppose a, b, and c are ints such that ab and
bc - Show ac or c a(int) remember ndk by
defn - ab ? b ar for some r?Z defn of div
- bc ? c bs for some s?Z defn of div
- c (ar)s substitution
- c a(rs) associative
- Let krs be an int mult of ints
- So, c ak
- ? a divides c or ac defn of div
4Proof by Counterexample
- Is it true or false that for all ints a and b, if
ab and ba then ab? - ?a,b ?Z, If ab and ba then ab
- Negating gives.
- ?x?Z (ab and ba) ? (a?b)
- Suppose a and b are ints such that ab and ba
- Show a?b
- ab ? b ak for some k?Z defn of div
- ba ? a bl for some l?Z defn of div
- b (bl)k b(lk) substitution assoc
- Since b ? 0, cancel bs giving 1lk (l and k are
divisors of 1) - Thus, k and l are both either 1 or -1
- If k l 1, then ba
- If k l -1, then b -a and so a?b
- This means you can find a counterexample by
taking b -a - Example, a -2 and b 2 ab 2-2 and ba
-22 but a?b - ? The proposed divisibility property is False!
5More Number Theory Definitions (cont.)
- Quotient-Remainder Theorem
- Given any integer n and positive int d
- ? unique q,r?Z ndq r and 0 r lt d
- Example 11/4
- 11 24 3
- Div/Mod
- n div d int quotient when n is divisible by d
- n mod d int remainder when n is divisible by d
- n div d q n mod d r
- n dq r
6Even/Odd is a Special Case of Divisibility
- We say that n is divisible by d if ?k?Z ndk
- n is divisible by 2 if ?k?Z n 2k (even)
- The other case is n 2k1 (odd, remainder of 1)
- n is divisible by 3 if ?k?Z n 3k
- The other cases are n 3k1 and n 3k2
- n is divisible by 4 if ?k?Z n 4k
- The other cases are n 4k1, n 4k2, n 4k3
- n is divisible by 5 if ?k?Z n 5k
- The other cases are .
7Parity of Integers
- How can we prove whether every integer is either
even or odd? - By Q-R Theorem we know that n dq r and 0 r
lt d - if d 2, then there exists integers q and r such
that - n 2q r and 0 r lt 2
- Evaluating the cases gives.
- n 2q 0 n 2q 1
- even parity (n2k) odd parity
(n2k1) - Theorem 3.4.2 Any two consecutive integers have
opposite parity
8Applying the Q-R Theorem
- Given any integer n, apply the Q-R Theorem to n
with d 4 - This implies that there exist an integer quotient
q and remainder r such that - n 4q r and 0 r lt 4
- Hence,
- n 4q, n 4q1, n 4q2, n 4q3
- Look at Theorem 3.4.3 The square of any odd
integer has the form 8m1
9Divisibility Proof
- Prove n2 2 is never divisible by 3 if n is an
integer - Discussion What does it mean for a number to be
divisible by 3? If a is divisible by 3 then ?k?Z
a3k and the remainder is 0. Other options are
a remainder of 1 and 2. So, we need to show that
the remainder when n2 2 is divided by 3 is
always 1 or 2. - There are 3 possible cases
- Case 1 n 3k
- Case 2 n 3k 1
- Case 3 n 3k 2
10n2-2 Proof (cont)
- Suppose n is a particular but arbitrarily chosen
integer - Show When n2 2 is divided by 3 the remainder
is always 1 or 2
Case 1 n 3k for ?k?Z n2-2 (3k)2 - 2
substitution 9k2 - 2 3(3k2) 2 mult
factoring 3(3k2 - 1) 1 rearranging ? The
remainder when dividing by 3 is 1 Case 2 n
3k1 for ?k?Z n2-2 (3k1)2 - 2
substitution 9k2 6k 1 - 2 3(3k2 2k)
1 mult factoring 3(3k2 2k - 1)
2 rearranging ? The remainder when dividing by
3 is 2
11n2-2 Proof (cont)
- Case 3 n 3k2 for ?k?Z
- n2-2 (3k2)2 - 2 substitution
- 9k2 12k 4 - 2 multiplying
- 3(3k2 4k) 2 rearranging
- ? The remainder when dividing by 3 is 2
- In each case the remainder when dividing n2-2 by
3 is nonzero. Thus proving the theorem.
12Unique Factorization Theorem
- Theorem 3.3.3 Any integer n gt 1 is either prime
or can be written as a product of prime numbers
in a way that is unique (Fundamental Theorem of
Arithmetic) - 1) prime number
- 2) product of prime numbers
- Example
- n 4 22, where 2 is a prime number
- n 7 71, where 7 is a prime number
- n 100 1010 2255, where 2 and 5 are
prime numbers
13Standard Factor Form
- Because of UFT, any integer n gt 1 can be written
in ascending order from left to right - n p1e1 p2e2 p3e3 . Pkek
- Where
- k is a positive integer
- p1 ? pk are prime numbers
- e1 ? ek are positive integers
- p1 lt p2 lt p3 lt lt pk
- Example
- k 100 1010 2255 2252
- n 3300 10033 425311 2255311 22
3 52 11
14Even More Number Theory Definitions (cont.)
- Floor/Ceiling
- Floor of x ? ? x ?
- unique integer n such that n x lt n1
- Ceiling of x ? ? x ?
- unique integer n such that n-1 lt x n
- Example
- X 37 / 4 9 ¼
- ? x ? 9 and 9 9 ¼ lt 10
- ? x ? 10 and 9 lt 9 ¼ 10
- Note Floor or Ceiling of an integer is itself!
15Proof by Counterexample
- Is the following True or False?
- ?x,y?R, ? x y? ? x ? ? y ?
- Method 1
- Suppose x and y are particular but arbitrarily
chosen real numbers such that x y ½ - Show The statement ? x y? ? x ? ? y ? is
False -
- ? ½ ½ ? ? 1 ? 1 substitution
- ? ½ ? ? ½ ? 0 0 0 substitution
- ? ? x y? ? ? x ? ? y ?
- Method 2
- Rewrite as a negation ? x y? ? ? x ? ? y ?
- Prove negation is True
16Other Floor/Ceiling Theorems
- Theorem 3.5.1 For all Real numbers x and all
integers m, ? x m ? ? x ? m - Intuitive since m is an integer and its floor is
always itself - Theorem 3.5.2 The Floor of n/2
- n/2 if n is even
- ? n/2 ?
- (n-1)/2 if n is odd
- Intuitive since when n is even, n/2 is an integer
and the floor of an integer is itself or n/2